Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $25.7$ years; the standard deviation is $2.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $21.5$ years.
Solution: $25.7$ $23.6$ $27.8$ $21.5$ $29.9$ $19.4$ $32$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $25.7$ years. We know the standard deviation is $2.1$ years, so one standard deviation below the mean is $23.6$ years and one standard deviation above the mean is $27.8$ years. Two standard deviations below the mean is $21.5$ years and two standard deviations above the mean is $29.9$ years. Three standard deviations below the mean is $19.4$ years and three standard deviations above the mean is $32$ years. We are interested in the probability of a tiger living less than $21.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the tigers will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $21.5$ years and the other half $({2.5\%})$ will live longer than $29.9$ years. The probability of a particular tiger living less than $21.5$ years is ${2.5\%}$.